The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $$\frac{h^2}{x \cdot m a_0^2}$$. The value of 10x is _________. ($$a_0$$ is radius of Bohr's orbit)
(Nearest integer)
[Given: $$\pi = 3.14$$]

For an electron moving in the $$n^{\text{th}}$$ Bohr orbit of a hydrogen atom, two standard relations are used repeatedly:

1. (Coulomb’s law equals centripetal force) $$\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}}$$

2. (Bohr’s quantisation of angular momentum) $$m v_{n} r_{n}=n\frac{h}{2\pi}$$

Eliminating $$v_{n}$$ from these two equations gives the well-known expression for the radius of the $$n^{\text{th}}$$ orbit,

$$r_{n}=n^{2}a_{0},$$

where the Bohr radius $$a_{0}$$ is

$$a_{0}= \frac{\varepsilon_{0}h^{2}}{\pi m e^{2}}.$$

We need the kinetic energy in the second orbit ($$n=2$$). From mechanics,

$$\text{K.E.}=\frac{1}{2}m v_{n}^{2}.$$

First we find $$v_{n}^{2}$$ from the force balance. Using the first relation,

$$\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}} \;\;\Longrightarrow\;\; v_{n}^{2}= \frac{e^{2}}{4\pi\varepsilon_{0}m r_{n}}.$$

Now substitute $$r_{n}=n^{2}a_{0}$$:

$$v_{n}^{2}= \frac{e^{2}}{4\pi\varepsilon_{0}m n^{2}a_{0}}.$$

Hence the kinetic energy is

$$\text{K.E.}= \frac12 m v_{n}^{2} =\frac12 m\left(\frac{e^{2}}{4\pi\varepsilon_{0}m n^{2}a_{0}}\right) =\frac{e^{2}}{8\pi\varepsilon_{0} n^{2} a_{0}}.$$

To convert the factor $$\dfrac{e^{2}}{4\pi\varepsilon_{0}}$$ into a form containing $$h^{2}$$ and $$a_{0}$$, we use the definition of $$a_{0}$$ written above. Starting with

$$a_{0}= \frac{\varepsilon_{0}h^{2}}{\pi m e^{2}},$$

rearrange to obtain

$$e^{2}= \frac{\varepsilon_{0}h^{2}}{\pi m a_{0}} \;\;\Longrightarrow\;\; \frac{e^{2}}{4\pi\varepsilon_{0}} =\frac{h^{2}}{4\pi^{2} m a_{0}}.$$

Insert this expression into the kinetic‐energy formula:

$$\text{K.E.}= \frac{1}{2 n^{2} a_{0}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) = \frac{1}{2 n^{2} a_{0}} \left(\frac{h^{2}}{4\pi^{2} m a_{0}}\right) =\frac{h^{2}}{8\pi^{2} n^{2} m a_{0}^{2}}.$$

For the second orbit, $$n=2\;$$ so $$n^{2}=4$$:

$$\text{K.E.}= \frac{h^{2}}{8\pi^{2}(4) m a_{0}^{2}} =\frac{h^{2}}{32\pi^{2} m a_{0}^{2}}.$$

This matches the required form $$\text{K.E.}= \frac{h^{2}}{x\,m a_{0}^{2}},$$ so

$$x = 32\pi^{2}.$$

Using $$\pi = 3.14$$ (as given),

$$\pi^{2}=3.14^{2}=9.8596,$$

$$x = 32 \times 9.8596 = 315.5072.$$

Now the question asks for $$10x$$ and its nearest integer value:

$$10x = 10 \times 315.5072 = 3155.072 \approx 3155.$$\

Hence, the correct answer is Option 3155.